Suppose $\{a_n\}$ is a sequence such that $\lim_{n\to\infty} (a_{n+1}-a_n)=0$.

Prove that if there exists $c>0$ such that for all $n$, $|a_n|\geq c$ then either for all but a finite number of terms $a_n>0$ or for all but a finite number of terms $a_n<0$.

I fail to see how $\lim_{n\to\infty} (a_{n+1}-a_n)=0$ helps here. I know that it means that for all $\varepsilon>0$, eventually $|a_{n+1}-a_n|<\varepsilon$. I tried playing with $\varepsilon$ by setting it to $c$ and use the triangle inequality but with no luck. Any tips?


HINT: If there are infinitely many "changes of signs" (i.e., if the conclusion doesn't hold), you can construct a subsequence of $\{a_{n+1}-a_n\}_n$ which doesn't converges to zero. Use the condition $|a_n|\geq c$.

The best way you can construct this sequence is this: Suppose $a_1$ is negative, let $a_{n_1+1}$ be the FIRST positive number in the sequence after $a_{1}$ (it exists by assumption), then let $a_{n_2+1}$ be the FIRST positive number in the sequence after $a_{n_1+1}$ and so on... argue that $a_{n_k}$ is negative for all $k$, $\{a_{n_k}\}$ will be your sequence. (A similar argument works if $a_1$ is positive).

  • $\begingroup$ I tried proving by contradiction. Can you elaborate please? What subsequence do you mean? Thanks for your reply. $\endgroup$ – user228929 Apr 5 '15 at 16:03
  • $\begingroup$ I claim that there exists a subsequence $\{a_{n_k}\}_k$ of $\{a_n\}$ such that $a_{n_k+1}a_{n_k}<0$ (i.e. $a_{n_k+1}$ and $a_{n_k}$ have opposite signs). Can you argue this? Once you have this, I'm sure you can prove that $|a_{n_k+1}-a_{n_k}|>2c$ for all $k$. (HINT: Suppose $a_{n_k+1}$ is possitive and $a_{n_k}$ is negative, then $a_{n_k+1}-a_{n_k}$ is a positive number so it's equal to its absolute value and $|a_{n_k+1}|=a_{n_k+1}$, $-a_{n_k}=|a_{n_k}|$). $\endgroup$ – Daniel Apr 5 '15 at 16:19
  • $\begingroup$ Let me verify: suppose the statement is false. Then there exist infinitely many positive elements of the sequence and infinitely many negative ones. So if $a_n>0$ for some $n$, I can find $N>n$ such that $a_N<0$. By taking such pairs of positive-negatives we get a subsequence $\{b_n\}$ of $\{a_n\}$. Then if we suppose $b_{n+1}>0$ and $b_n<0$, $|b_{n+1}-b_{n}|=|b_{n+1}|+|b_{n}|\geq 2c$. That's more or less what I did. But the initial condition with limit applies to $\{a_n\}$ and not $\{b_n\}$. Where am I wrong? $\endgroup$ – user228929 Apr 5 '15 at 16:59
  • $\begingroup$ You're right until this point: "If $a_n>0$, I can find $N>n$ such that $a_N<0$. I don't know if it's clear but there is must be an $i$ between $n$ and $N$ (inclusive) such that $a_i$ is positive and $a_{i+1}$ is negative (there must be a change of sign somewhere). Your final argumentis fine. $\endgroup$ – Daniel Apr 5 '15 at 17:05
  • $\begingroup$ I edited the answer. Just in case. $\endgroup$ – Daniel Apr 5 '15 at 17:10

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